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Posers and Puzzles

Posers and Puzzles

  1. Standard member PBE6
    Bananarama
    16 Oct '08 16:03
    That's the question! I think this one was actually posed in the body of another thread, but let's make the request formal:

    If you ask a group of "n" people "how many people in this group will answer this question correctly?", what is the probability that the question will be answered correctly by "m" people?

    The question itself is a little confusing, so I'll provide a few examples. If there are 2 people in the group, here are the possible responses:

    (0,0)
    (1,0)
    (0,1)
    (1,1)
    (2,0)
    (0,2)
    (2,1)
    (1,2)
    (2,2)

    If both answer "2", i.e. (2,2), then both people are correct. If either player answers "1" and the other does not, i.e. (1,0), (1,2), (0,1), (2,1), then the player that answered "1" is correct. If both players answer "0", i.e. (0,0), then the result is undecidable because of the following logic loop: neither player answered "2", which means both were wrong, which means the correct answer is "0", but both players answered "0" which means that both players were right, but neither player actually answered "2", which means that... (etc...). Similar undecidable situations arise with the other answers, and in larger groups where there are multiple correct answer clusters (which, funnily enough, means that 2 rights make a wrong!).

    Hopefully the question makes sense, and we come up with an answer that also makes sense. Enjoy!
  2. Standard member Palynka
    Upward Spiral
    16 Oct '08 16:29 / 2 edits
    Originally posted by PBE6
    That's the question! I think this one was actually posed in the body of another thread, but let's make the request formal:

    [b]If you ask a group of "n" people "how many people in this group will answer this question correctly?", what is the probability that the question will be answered correctly by "m" people?


    The question itself is a little confusi akes sense, and we come up with an answer that also makes sense. Enjoy![/b]
    Is there any gain in being the sole correct answer? Specifically, if I win because I say 1 and the other one says 2, do I win more than if both of us say 2?
  3. Standard member PBE6
    Bananarama
    16 Oct '08 17:02 / 1 edit
    Originally posted by Palynka
    Is there any gain in being the sole correct answer? Specifically, if I win because I say 1 and the other one says 2, do I win more than if both of us say 2?
    Nope. We can make that a follow-up question though. I'm just trying to see what the probability of this question being answered correctly is for any given correct answer.

    Did a teeny-tiny bit of work on my own. In the original example, the following answer sets result in no winners:

    (0,0), (0,2), (2,0), (1,1)

    The following answer sets result in one or more winners:

    (1,0), (0,1), (1,2), (2,1), (2,2)

    The bold numbers above represent the winner(s). Of these answers, 4 were won with the number "1", and 1 was won with the number "2", so the overall probability of "1" being the winning number is 4/9 and the overall probability of "2" being the winning number is 1/9. I want to extend this to any "m" within a group of "n" people, but it starts to look a bit complicated the bigger the groups get.
  4. 16 Oct '08 17:15
    Originally posted by PBE6
    That's the question! I think this one was actually posed in the body of another thread, but let's make the request formal:

    [b]If you ask a group of "n" people "how many people in this group will answer this question correctly?", what is the probability that the question will be answered correctly by "m" people?


    The question itself is a little confusi ...[text shortened]... akes sense, and we come up with an answer that also makes sense. Enjoy![/b]
    You wrote:
    If either player answers "1" and the other does not, i.e. (1,0), (1,2), (0,1), (2,1), then the player that answered "1" is correct.

    I do not understand why. maybe in the case (1,2) nobody is correct - the answer is 0 and no one said so?
  5. Standard member PBE6
    Bananarama
    16 Oct '08 17:30
    Originally posted by David113
    You wrote:
    If either player answers "1" and the other does not, i.e. (1,0), (1,2), (0,1), (2,1), then the player that answered "1" is correct.

    I do not understand why. maybe in the case (1,2) nobody is correct - the answer is 0 and no one said so?
    In my original conception, the answer is correct if m people in the group pick the number "m" (i.e., 1 person picks 1, 2 people pick 2, 3 people pick 3, etc...), just because it logically seemed to make sense. However, now that you've asked the question, I think I can justify it as part of an algorithm.

    Say the answers given were (1,2). Let's assume the "correct" answer is supposed to be "2". If that's the case, then 1 person got it right. Then the answer to the question "how many people will answer this question correctly?" is 1. If that's the case, then "1" is really the right answer, not "2" as assumed in the first place.

    "0" is a bit trickier. Let's assume the "correct" answer is supposed to be "0". If that's the case, then 0 people got it right. Then the answer to the question "how many people will answer this question correctly?" is 0. No problem there. However, this answer works in every possible case (i.e. (1,2,3), no one said "0" so "0" is the right answer here too, etc...), so it's reasonable to consider this a degenerate case.

    In the case of multiple correct answer clusters (i.e. (2,2,3,3,3), 2 people said "2" so that's right, but 3 people said "3" so that's right too, etc...), we have several "correct" answers to the same question, which in my mind makes the question undecidable with the given answer set.

    Feel free to remark on the logic here, it's a bit loopy (a word which has recently re-entered my vocabulary because I'm reading "I Am A Strange Loop" by Douglas Hofstadter, which not-so-coincidentally discusses Godel's Incompleteness Theorem and the strange loopiness of all mathematics!).
  6. Standard member Palynka
    Upward Spiral
    16 Oct '08 17:34 / 1 edit
    Originally posted by PBE6
    Nope. We can make that a follow-up question though. I'm just trying to see what the probability of this question being answered correctly is for any given correct answer.

    Did a teeny-tiny bit of work on my own. In the original example, the following answer sets result in no winners:

    (0,0), (0,2), (2,0), (1,1)

    The following answer sets result in one p of "n" people, but it starts to look a bit complicated the bigger the groups get.
    It seems to me that the only Nash equilibrium is when everybody answers N (where N is the total number of players). Only then nobody has an incentive to change strategy.
  7. Standard member PBE6
    Bananarama
    16 Oct '08 17:43 / 3 edits
    Originally posted by Palynka
    It seems to me that the only Nash equilibrium is when everybody answers N (where N is the total number of players). Only then nobody has an incentive to change strategy.
    I never thought about it terms of game theory, but that's a pretty interesting answer about which number is the best to pick. I just thought it was a funny question to ask a group of unsuspecting party goers.

    As a follow-up question, in addition to the original question (an answer to which is still pending), what if we change this into a game where everybody pays $1 to play, the house keeps $0.50 as a rake, and then the (n-0.5) dollars are split up amongst the winning players? For instance, if the correct answer is 5 in a game with 10 players, each winner gets ($10-$0.50)/5 = $9.50/5 = $1.90. In this game, if everyone guess 10, then each player gets ($10-$0.50)/10 = $9.50/10 = $0.95, which is a losing proposition.

    EDIT: Numbers and stupid unexpected smileys...
  8. Standard member Palynka
    Upward Spiral
    16 Oct '08 18:04
    Originally posted by PBE6
    In my original conception, the answer is correct if m people in the group pick the number "m" (i.e., 1 person picks 1, 2 people pick 2, 3 people pick 3, etc...), just because it logically seemed to make sense. However, now that you've asked the question, I think I can justify it as part of an algorithm.

    Say the answers given were (1,2). Let's assume the "c ...[text shortened]... orem and the strange loopiness of all mathematics!).
    I think the only coherent answer to those issues is to remove 0 from the choice set (even if we found a coherent way to be able to include it, nobody could ever win with it!) and to consider (2,2,3,3,3) as 0 winners.
  9. Standard member uzless
    The So Fist
    16 Oct '08 18:05
    Originally posted by PBE6
    That's the question! I think this one was actually posed in the body of another thread, but let's make the request formal:

    [b]If you ask a group of "n" people "how many people in this group will answer this question correctly?", what is the probability that the question will be answered correctly by "m" people?


    The question itself is a little confusi ...[text shortened]... akes sense, and we come up with an answer that also makes sense. Enjoy![/b]
    If you assume everyone will pick a random number, limited only by how many are the in the group, isn't this just like that birthday game where you ask what is the minimum number of people in a room that will have the same birth date?
  10. Standard member Palynka
    Upward Spiral
    16 Oct '08 18:06
    Originally posted by PBE6
    I never thought about it terms of game theory, but that's a pretty interesting answer about which number is the best to pick. I just thought it was a funny question to ask a group of unsuspecting party goers.

    As a follow-up question, in addition to the original question (an answer to which is still pending), what if we change this into a game where every ...[text shortened]... 50/10 = $0.95, which is a losing proposition.

    EDIT: Numbers and stupid unexpected smileys...
    Without game theory, I don't know what the question is supposed to be!

    I'm feeling a bit lazy to work on the other one...
  11. Standard member PBE6
    Bananarama
    16 Oct '08 18:08
    Originally posted by Palynka
    I think the only coherent answer to those issues is to remove 0 from the choice set (even if we found a coherent way to be able to include it, nobody could ever win with it!) and to consider (2,2,3,3,3) as 0 winners.
    I agree.
  12. 16 Oct '08 18:25
    Originally posted by PBE6
    Nope. We can make that a follow-up question though. I'm just trying to see what the probability of this question being answered correctly is for any given correct answer.

    Did a teeny-tiny bit of work on my own. In the original example, the following answer sets result in no winners:

    (0,0), (0,2), (2,0), (1,1)

    The following answer sets result in one ...[text shortened]... p of "n" people, but it starts to look a bit complicated the bigger the groups get.
    And '0' always looses? Why would anyone say 0 then?
  13. 16 Oct '08 22:33 / 1 edit
    Supposing you ask 6 (or more) people this question. And suppose of these, you have 1 person answer "1", 2 people answer "2", and 3 people answer "3".

    Which of the answers is correct?

    The problem with the question is you have to know the correct answer to know how many got it right, and you have to know how many got it right to know the correct answer.

    Another example.. Supposing you ask 6 people the question. The answers you receive are as follows (2,1,2,1,3,2). How many answered correctly?

    It does make for an interesting poser, though, as it does introduce the concept of differentiating a question from a metaquestion.
  14. Standard member Palynka
    Upward Spiral
    17 Oct '08 08:29
    Originally posted by geepamoogle
    Supposing you ask 6 (or more) people this question. And suppose of these, you have 1 person answer "1", 2 people answer "2", and 3 people answer "3".

    Which of the answers is correct?

    The problem with the question is you have to know the correct answer to know how many got it right, and you have to know how many got it right to know the correct ans ...[text shortened]... hough, as it does introduce the concept of differentiating a question from a metaquestion.
    See my posts above.
  15. Standard member forkedknight
    Defend the Universe
    23 Oct '08 15:56
    Originally posted by geepamoogle
    Supposing you ask 6 (or more) people this question. And suppose of these, you have 1 person answer "1", 2 people answer "2", and 3 people answer "3".

    Which of the answers is correct?

    The problem with the question is you have to know the correct answer to know how many got it right, and you have to know how many got it right to know the correct ans ...[text shortened]... hough, as it does introduce the concept of differentiating a question from a metaquestion.
    The question would make much more sense if it was stated, "How many people will choose the same answer as you?"

    Then,

    m is a correct answer if m people choose m.

    This results in almost the same result map as PEB6 originally stated, except the winners are more clear.

    Supposing you ask 6 (or more) people this question. And suppose of these, you have 1 person answer "1", 2 people answer "2", and 3 people answer "3".

    Which of the answers is correct?


    In this case, all people would be correct.

    Another example.. Supposing you ask 6 people the question. The answers you receive are as follows (2,1,2,1,3,2). How many answered correctly?

    In this case, no one would be correct.